Metamath Proof Explorer


Theorem perpln2

Description: Derive a line from perpendicularity. (Contributed by Thierry Arnoux, 27-Nov-2019)

Ref Expression
Hypotheses perpln.l ⊢ L = Line 𝒢 ⁡ G
perpln.1 ⊢ φ → G ∈ 𝒢 Tarski
perpln.2 ⊢ φ → A ⟂ 𝒢 ⁡ G B
Assertion perpln2 ⊢ φ → B ∈ ran ⁡ L

Proof

Step Hyp Ref Expression
1 perpln.l ⊢ L = Line 𝒢 ⁡ G
2 perpln.1 ⊢ φ → G ∈ 𝒢 Tarski
3 perpln.2 ⊢ φ → A ⟂ 𝒢 ⁡ G B
4 df-perpg ⊢ ⟂ 𝒢 = g ∈ V ⟼ a b | a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g
5 simpr ⊢ φ ∧ g = G → g = G
6 5 fveq2d ⊢ φ ∧ g = G → Line 𝒢 ⁡ g = Line 𝒢 ⁡ G
7 6 1 eqtr4di ⊢ φ ∧ g = G → Line 𝒢 ⁡ g = L
8 7 rneqd ⊢ φ ∧ g = G → ran ⁡ Line 𝒢 ⁡ g = ran ⁡ L
9 8 eleq2d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ↔ a ∈ ran ⁡ L
10 8 eleq2d ⊢ φ ∧ g = G → b ∈ ran ⁡ Line 𝒢 ⁡ g ↔ b ∈ ran ⁡ L
11 9 10 anbi12d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ↔ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L
12 5 fveq2d ⊢ φ ∧ g = G → ∟ 𝒢 ⁡ g = ∟ 𝒢 ⁡ G
13 12 eleq2d ⊢ φ ∧ g = G → ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
14 13 ralbidv ⊢ φ ∧ g = G → ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
15 14 rexralbidv ⊢ φ ∧ g = G → ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
16 11 15 anbi12d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
17 16 opabbidv ⊢ φ ∧ g = G → a b | a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
18 2 elexd ⊢ φ → G ∈ V
19 1 fvexi ⊢ L ∈ V
20 rnexg ⊢ L ∈ V → ran ⁡ L ∈ V
21 19 20 mp1i ⊢ φ → ran ⁡ L ∈ V
22 21 21 xpexd ⊢ φ → ran ⁡ L × ran ⁡ L ∈ V
23 opabssxp ⊢ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ L × ran ⁡ L
24 23 a1i ⊢ φ → a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ L × ran ⁡ L
25 22 24 ssexd ⊢ φ → a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ∈ V
26 4 17 18 25 fvmptd2 ⊢ φ → ⟂ 𝒢 ⁡ G = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
27 26 rneqd ⊢ φ → ran ⁡ ⟂ 𝒢 ⁡ G = ran ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
28 23 rnssi ⊢ ran ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ ran ⁡ L × ran ⁡ L
29 27 28 eqsstrdi ⊢ φ → ran ⁡ ⟂ 𝒢 ⁡ G ⊆ ran ⁡ ran ⁡ L × ran ⁡ L
30 rnxpss ⊢ ran ⁡ ran ⁡ L × ran ⁡ L ⊆ ran ⁡ L
31 29 30 sstrdi ⊢ φ → ran ⁡ ⟂ 𝒢 ⁡ G ⊆ ran ⁡ L
32 relopabv ⊢ Rel ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
33 26 releqd ⊢ φ → Rel ⁡ ⟂ 𝒢 ⁡ G ↔ Rel ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
34 32 33 mpbiri ⊢ φ → Rel ⁡ ⟂ 𝒢 ⁡ G
35 brrelex12 ⊢ Rel ⁡ ⟂ 𝒢 ⁡ G ∧ A ⟂ 𝒢 ⁡ G B → A ∈ V ∧ B ∈ V
36 34 3 35 syl2anc ⊢ φ → A ∈ V ∧ B ∈ V
37 36 simpld ⊢ φ → A ∈ V
38 36 simprd ⊢ φ → B ∈ V
39 brelrng ⊢ A ∈ V ∧ B ∈ V ∧ A ⟂ 𝒢 ⁡ G B → B ∈ ran ⁡ ⟂ 𝒢 ⁡ G
40 37 38 3 39 syl3anc ⊢ φ → B ∈ ran ⁡ ⟂ 𝒢 ⁡ G
41 31 40 sseldd ⊢ φ → B ∈ ran ⁡ L